Why is barium sulfate insoluble in water
EL-MO I elements, textbook
Equal-ionic additive - Precipitations KM-6: Equilibrium - transfer 119 119 Equal-ionic additive The solubility of salts can be reduced by introducing a type of ion present in the salt - equal-ionic additive (influencing the equilibrium). Barium sulfate is used as a contrast medium for gastric x-rays. Barium ions are harmful to the organism. In order to keep the barium ion concentration as low as possible, a slurry of poorly soluble barium sulphate is prepared in a sodium sulphate solution (sodium sulphate is readily soluble). Example Calculation of the solubility of barium sulfate: K SP = 10 -10 a) in 1 liter of water BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) x = √ K SP = 10 -5 mol / L ⇒ S (BaSO 4) = 233.4 • 10 -5 = 2.33 • 10 -3 g / L b) in 1 liter of sodium sulfate solution with c = 0.1 mol / liter of BaSO 4 (s) Ba 2+ (aq) + SO 4 2– (aq) at the beginning: certain amount 0 0.1 (from the Na 2 SO 4 solution) in the equation: certain amount - xx 0.1 + x K SP = x • ( 0.1 + x) This quadratic equation is solvable, but the process can be simplified. In pure water only 10 -5 mol sulfate ions were formed; the sulphate ion concentration from the barium sulphate is suppressed by the addition of the same ions. For this reason, x can be neglected in the sum compared to 0.1. The above relationship is therefore simplified to: K SP = 10 -10 = 0.1 • x ⇒ x = 10 -9 mol / L m (BaSO 4) in 1 liter of Na 2 SO 4 solution (c = 0.1 mol / Liter) = 2.33 • 10 -7 g / L. Precipitation reactions When ion solutions of easily soluble salts are poured together, ion combinations of a poorly soluble salt can be formed (Fig. 119–1). If the solubility product of the salt is exceeded, a precipitate forms (“precipitation”). Precipitation reactions play a major role in wastewater treatment. Harmful ions (eg phosphates) can be bound as a separable precipitate by adding salts (eg FeCl 3). In the qualitative analysis of salt solutions, the first step is usually a precipitation reaction. All silver halides are sparingly soluble. If halides are suspected in a solution, the addition of a silver nitrate solution (AgNO 3 is easily soluble) can be used to corroborate the assumption that a precipitate has formed. In order to obtain unambiguous statements, further steps have to be followed, since there are other poorly soluble silver salts (Fig. 119–2). Fig. 119–1: Formation of a precipitate Solution of salt A + D - Soluble salt C + D - Soluble salt A + B - Insoluble salt C + B - 3 samples A, B and C may contain iodide ions. All 3 samples become mixed with silver nitrate Possible reaction: I - + Ag + AgI Result: Only samples B and C contain iodide ions. Exercise 119.1 Solubility Calculate the solubility of silver bromide in 1 liter of water or in 1 liter of NaBr solution c = 0 , 1 mol / liter! Student experiment 4.2 Precipitation reactions Student experiment 4.1 Equal ionic additive 10 mL NaCl solution (c = 0.2 mol / L) and 10 mL AgNO 3 solution (c = 0.2 mol / L) are mixed. Does AgCl precipitate (K SP = 2 • 10 –10)? Solution 20 mL: c (Ag +) = 0.2 / 2 = 0.1 mol / L c (Cl -) = 0.2 / 2 = 0.1 mol / L c (Ag +) • c (Cl - ) = 0.01 0.01> 2 • 10 –10 c Therefore a precipitate of AgCl forms! VS VS Fig. 119–2: Detection of iodide ions through the formation of precipitation For testing purposes only - property of the publisher öbv
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